ach linear equation given (ii) y = y = x – (iv) y = 4x – 5 (vi) 2x – 3y = 4 plz give me equation it’s u

Question

ach linear equation given
(ii) y =
y = x –
(iv) y = 4x –
5
(vi) 2x – 3y = 4

plz give me equation it’s urgent​

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Maya 7 hours 2021-11-25T04:12:36+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-25T04:14:15+00:00

    Answer:

    (i) x + y =5 and 2x –3y = 4

    By elimination method

    x + y =5 … (i)

    2x –3y = 4 … (ii)

    Multiplying equation (i) by (ii), we get

    2x + 2y = 10 … (iii)

    2x –3y = 4 … (ii)

    Subtracting equation (ii) from equation (iii), we get

    5y = 6

    y = 6/5

    Putting the value in equation (i), we get

    x = 5 – (6/5) = 19/5

    Hence, x = 19/5 and y = 6/5

    By substitution methodx + y = 5 … (i)

    Subtracting y both side, we get

    x = 5 – y … (iv)

    Putting the value of x in equation (ii) we get

    2(5 – y) – 3y = 4

    -5y = – 6

    y = -6/-5 = 6/5

    Putting the value of y in equation (iv) we get

    x = 5 – 6/5

    x = 19/5

    Hence, x = 19/5 and y = 6/5 again

    (ii) 3x + 4y = 10 and 2x – 2y = 2

    By elimination method

    3x + 4y = 10 …. (i)

    2x – 2y = 2 … (ii)

    Multiplying equation (ii) by 2, we get

    4x – 4y = 4 … (iii)

    3x + 4y = 10 … (i)

    Adding equation (i) and (iii), we get

    7x + 0 = 14

    Dividing both side by 7, we get

    x = 14/7 = 2

    Putting in equation (i), we get

    3x + 4y = 10

    3(2) + 4y = 10

    6 + 4y = 10

    4y = 10 – 6

    4y = 4

    y = 4/4 = 1

    Hence, answer is x = 2, y = 1

    By substitution method

    3x + 4y = 10 … (i)

    Subtract 3x both side, we get

    4y = 10 – 3x

    Divide by 4 we get

    y = (10 – 3x )/4

    Putting this value in equation (ii), we get

    2x – 2y = 2 … (i)

    2x – 2(10 – 3x )/4) = 2

    Multiply by 4 we get

    8x – 2(10 – 3x) = 8

    8x – 20 + 6x = 8

    14x = 28

    x = 28/14 = 2

    y = (10 – 3x)/4

    y = 4/4 = 1

    Hence, answer is x = 2, y = 1 again.

    (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

    By elimination method

    3x – 5y – 4 = 0

    3x – 5y = 4 …(i)

    9x = 2y + 7

    9x – 2y = 7 … (ii)

    Multiplying equation (i) by 3, we get

    9 x – 15 y = 11 … (iii)

    9x – 2y = 7 … (ii)

    Subtracting equation (ii) from equation (iii), we get

    -13y = 5

    y = -5/13

    Putting value in equation (i), we get

    3x – 5y = 4 … (i)

    3x – 5(-5/13) = 4

    Multiplying by 13 we get

    39x + 25 = 52

    39x = 27

    x =27/39 = 9/13

    Hence our answer is x = 9/13 and y = – 5/13

    By substitution method

    3x – 5y = 4 … (i)

    Adding 5y both side we get

    3x = 4 + 5y

    Dividing by 3 we get

    x = (4 + 5y )/3 … (iv)

    Putting this value in equation (ii) we get

    9x – 2y = 7 … (ii)

    9 ((4 + 5y )/3) – 2y = 7

    Solve it we get

    3(4 + 5y ) – 2y = 7

    12 + 15y – 2y = 7

    13y = – 5

    y = -5/13

    x = 4 + 5 ( -5/13)/ 3

    = 4 – 25/13 / 3

    = 4 × 13 – 25/13 / 3

    = 27/13×3

    = 27/39

    = 9/13

    Hence we get x = 9/13 and y = – 5/13 again.

    (iv) x/2 + 2y/3 = – 1 and x – y/3 = 3

    By elimination method

    x/2 + 2y/3 = -1 … (i)

    x – y/3 = 3 … (ii)

    Multiplying equation (i) by 2, we get

    x + 4y/3 = – 2 … (iii)

    x – y/3 = 3 … (ii)

    Subtracting equation (ii) from equation (iii), we get

    5y/3 = -5

    Dividing by 5 and multiplying by 3, we get

    y = -15/5

    y = – 3

    Putting this value in equation (ii), we get

    x – y/3 = 3 … (ii)

    x – (-3)/3 = 3

    x + 1 = 3

    x = 2

    Hence our answer is x = 2 and y = −3.

    By substitution method

    x – y/3 = 3 … (ii)

    Add y/3 both side, we get

    x = 3 + y/3 … (iv)

    Putting this value in equation (i) we get

    x/2 + 2y/3 = – 1 … (i)

    (3+ y/3)/2 + 2y/3 = -1

    3/2 + y/6 + 2y/3 = – 1

    Multiplying by 6, we get

    9 + y + 4y = – 6

    5y = -15

    y = – 3

    Hence our answer is x = 2 and y = −3.

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